SPOJ VLATTICE Visible Lattice Points 莫比乌斯反演 难度:3

http://www.spoj.com/problems/VLATTICE/





明显,当gcd(x,y,z)=k,k!=1时,(x,y,z)被(x/k,y/k,z/k)遮挡,所以这道题要求的是gcd(x,y,z)==1的个数+{(x,y,0)|gcd(x,y)==1}的个数+3{(0,0,1),(0,1,0),(1,0,0)}

现在不去管最后的三个坐标轴上的点,

设f(i)=|{(x,y,0)|gcd(x,y)==i}|*3+|{(x,y,z)|gcd(x,y,z)==i}|,也就是不在坐标轴上且非0坐标值的最大公约数为n的个数,

设F(i)为由能被i整除的坐标值组成的不在坐标轴上的坐标的个数,则F(i)=n/i*n/i*(n/i+3),同时显然F(i)=sigma(b|n,f[i]),

由莫比乌斯反演,可得f(1)=sigma(mul(i)*F(i))

#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn =1e6+2;
bool ifprime[maxn];
int mul[maxn];
int prime[maxn];
void moblus(){
ifprime[2]=true;
for(int i=3;i<maxn;i+=2)ifprime[i]=true;
int pnum=0;
mul[1]=1;
for(int i=2;i<maxn;i++){
if(ifprime[i]){
prime[pnum++]=i;
mul[i]=-1;
}
for(int j=0;j<pnum&&i*prime[j]<maxn;j++){
ifprime[i*prime[j]]=false;
if(i%prime[j]==0){
mul[i*prime[j]]=0;
break;
}
else {
mul[i*prime[j]]=-mul[i];
}
}
}
}
int main(){
int T;
moblus();
scanf("%d",&T);
while(T--){
int n;
scanf("%d",&n);
long long ans=3;
for(int i=1;i<=n;i++){
ans+=(long long)mul[i]*(n/i)*(n/i)*(n/i+3);
}
printf("%I64d\n",ans);
}
return 0;
}